Question

A hydraulic lift at a service station can lift cars with a mass of \[3500\,kg\]. The area of the cross section of the piston carrying the load is \[500\,c{m^2}\]. What pressure does the smaller piston experience? Take \[g = 9.8\,m/{s^2}\]?

Answer 1

Hint: The pressure due to applied force is the ratio of magnitude of force and the area of the cross section on which the force is applied.
\[{\text{pressure = }}\dfrac{{{\text{force}}}}{{{\text{area}}\,{\text{of}}\,{\text{cross}}\,{\text{section}}}}\]

Complete step by step answer:
According to Newton’s law of gravitation, the gravitational force exerted on the car of mass m is given by the equation,
\[F = mg\]

Here, g is acceleration due to gravity.

The pressure exerted on the smaller piston of the hydraulic lift is the ratio of force exerted on the car to the area of the cross section of the smaller piston.

Therefore,
\[P = \dfrac{F}{A}\]
\[ \Rightarrow P = \dfrac{{mg}}{A}\]

Substitute \[3500\,kg\] for m, \[9.8\,m/{s^2}\] for g and \[500\,c{m^2}\] for A in the above equation.
\[P = \dfrac{{\left( {3500\,kg} \right)\left( {9.8\,m/{s^2}} \right)}}{{\left( {500\,c{m^2}} \right)\left( {\dfrac{{{{10}^{ - 4}}\,{m^2}}}{{1\,c{m^2}}}} \right)}}\]
\[ \Rightarrow P = \dfrac{{34300\,kg\,m/{s^2}}}{{500 \times {{10}^{ - 4}}\,{m^2}}}\,\]
\[\therefore P = 6.86 \times {10^5}\,N/{m^2}\]

Therefore, the pressure experienced by the smaller piston is \[6.86 \times {10^5}\,N/{m^2}\].

Note:
The weight of the object is the gravitational force exerted on the object. Sometimes students misunderstand between the mass and the weight of the object. 3500 kg is the mass of the car and not the weight. Therefore, you need to calculate the weight of the car by multiplying its mass by the acceleration due to gravity. Also, \[1\,cm = {10^{ - 2}}\,m\], therefore, \[1\,c{m^2} = {\left( {{{10}^{ - 2}}\,m} \right)^2} = {10^{ - 4}}\,{m^2}\].